HAHAHAHA....Damned foreign languages hehehehe
Bwild....you a buggar of a man:)
Fencer....do you have a criminal record? LOLOL prolly not but i bet there are a few here that do ROFLMBO
AD...I would love to see more of you BUT....and that is a big butt hehehe your butt I mean ROFLMBO

again sorry Harley :)....blame da Boss

chattytea: They do it through editing their details but this change must be approved by me or Liquid.
Bernice: Show us your real photo first :-)

I dont want to make a pest of myself with changing pics...my "eye" was disturbing to some....no worries... my dog is bringing some hilarious remarks hehehehehe.....my real pic.??
If yours can be 3 years old dear Fencer, can mine be 20 years old...Im on the decline side you see and dont need smart remarks from some of your members..no names but you all know where im coming from LOLOL

Im trying to make my new fellowships into my "favourites" :(
I cant...has something changed or do i have to keep clicking on my "Fellowships" list??

the fellowships dont go into the favourites nor bernice.. any new messages show up by the side of any fellowship, and just click on that red number to read the message.

Thank you fencer, I will tell them about that, I thought things had changed now :)

thanks Chatty, but it makes for extra "clicks"

I know...dont be so lazy.. but i can see on the bottom half of my main page all my fellowships, having to go there and then click on number of posts isnt an incentive to do that exactly ( but then that depends on the fellowship i guess) there are some fiery buggars LOLOL

A photo of you taken 20 years ago, Bernice? You would have been a toddler then, surely? :o)

No,she was young and pretty at this time..:)

hahahah 20 years ago....I was about your age bumble .....40...hehehhe
**edit** hehehe and here come the remarks....thanks Bumble, we both know what it is like to decline in our later years dont we :)

I can't remember what it was like to decline in my later years. :o)

9 years ago I was my age as well!

Yes its true Bernice. and I was never told if the other party done thier bit in the end.

Kevin, at your age you should be out and about anyways. Not playing silly games with other peoples time out limits to annoy them.

I think thats more than enough about Kevin. He's playing within the rules, and is a lovely person too. Theres no need to single him out, he's not the only person to take his time at games, I have been myself lately because I've been busy.

Backoff has made me big boss of his tablut fellowship but I'm not getting "define new tournament" in the team box, any advice?

That's to set up a fellowship tournament for that game type.

Yes it seems most people are using their own picutres. Mine hame is certainly an original. There are some that crack me up though like AD's. Too funny AD. I think I saw Pres Clintons out there somewhere too or I might be mistaken.

ughaibu: It will be a server-side cache problem [I've written the cache myself and it contains some minor bugs]. The application server reload should fix it, I'll ask Liquid to do it.

It should be done.

but I really do look like this :)

town tell wednesday. Sry if i did not get to our game..
:o{P

thank you, everything worked out perfectly:)

...someone could create the "Back In The Day" Fellowship where members could upload everything from their baby pictures to when they were teenages, to young adults, then old cronies :)

Yours was taken in your moody period then? ;oÞ

Actually, it was part of this picture where my daughter and I both had the same exact hand in poker. I was getting ready to play in a Texas Hold'Em tournament in Australia, so I was using the Adelaide Casino deck (blue diamonds and green clubs.)

Brittany likes to play 5-card draw so I was taking a break playing a few hands with her.

We both ended up with a pair of tens, and an Ace-Queen-2. It was kind of funny, because at the show down she said "Pair of tens". So I asked her what her next high card was. She said "I have an Ace, so I win." I laughed and said I had an Ace too. I told her I had a Queen, then she laughed and said "darn, you probably will win, I only have a 2."

It was such an unlikely occurence, one of the other players took a picture of it for us.

Wow ! what are the chances ??
don't worry, that was a rhetorical question ;-D

Oh don't, Niki! He'll give us a huge mathematical formula as to exactly what the chances are! (Only kidding, GI!)

Fencer, just checked my rating in my profile and it says i am 3rd on BB+ , but when you click on it and goto the ratings it says 2nd.

Actually harls, I did start to! But, I realized I was making an error since the fact we both had a pair changes the math considerably.

MM: See http://brainking.info/bugs.html.

At least you have a rank - in Amazons, I have a rating, but no rank! :-(

(Bug where rank is not updated if rating does not change, and my rating changed from "unrated" to XXXX - but moved "0" and hence not really "updated", hence not given a rank) :-)

That doesn't surprise me! And it probably wouldn't surprise you that I wouldn't have understood a number of it! :oD

ok
I had not bothered to look as the date still says 30th June lol :-)
No wonder i keep jumping about on the ranking at the moment, mainly against unrated players.

The chances of the two players getting the same pair and the same three (different) other cards is:

(52/52 * 3/51 * 48/50 * 44/49 * 40/48) * (2/47 * 1/46 * 3/45 * 3/44 * 3/43) = 1.24 * 10 ^ -8 = 0.0000000124 = 0.000000124 % = 124 / 10,000,000,000 = 1 / 80,667,372.92

*faint* Amazing odds, however you worked that out!

rofl - It's amazing what you can do with a calculator :-)

and it only took Kevin 44 minutes to do as well, :)

Except that the answer is wrong.

There are 13 of the same cards, disregarding suits.


The chances of 4 of the same being drawn by two people would be...

4 x (13x12x11x10/4x3x2) then divided by 2.

This is one in 5720.

Now, given that 4 cards have been removed, with 6 more to be dealt, with 3 matching pairs, the odds are:

{there are just 12 "different" cards now to start...}

4 x (12 x 12) x (11 x 11) x (10 x 10) =

6,969,600.

Multiply this by 5720 and you get:


39,866,112,000 to 1.

So then what is the answer, if my answer is wrong?

I believe your answer is not correct either.

You say:
--------
{there are just 12 "different" cards now to start...}

4 x (12 x 12) x (11 x 11) x (10 x 10) =

6,969,600.
-----
However, after the first "x12", it is no longer 12:1 that another card of the same number will be drawn - there are only three of it remaining, while there are four of the rest. Same for the 2nd "x11" and "x10".

go play your games ;oÞ

No more Stevie or I'll have to delete. Not even with a smiley face!

poohy LOL

Kevin, if you have 4 groups of 13 different cards, then pull out all 4 10's, there are just 12 different ones remaining, correct?

2 2 2 2
3 3 3 3
4 4 4 4
. . . .
A A A A

OK, now I am going to remove 1, then remove another exactly like it.

Surely there are 12 sets of matching cards to start with. And, even after I draw 1 card, there are still 12 identical sets. One set will only have 3 cards, but that does not matter.

Now, there are 11 of the same sets with 4 cards, and one set of cards only has 2.

I know I am not drawing from the set that has 2, or I will have another pair.

That means I am drawing from one of the 11.

Etc.

If you know you are not drawing from the group that has two, then you have not accounted for that in your math. Also, it does matter whether the group has four or the group has three. The chances of selecting each individual card is the same (1 / (# of cards)) - not the chance of getting a 9 is the same as getting a Queen, regardless of how many of each are left.

Harley, this is off topic for the board????

Sorry Kevin I am with Ed. LOL

Yes I have.

The odds of drawing 1 out of 12 if 1 out of 12. It does not matter is I have all 4 suits, or 3 of the 4 suits, or just 1 of the four suits.

The odds of picking 1 out of 12 is 1 out of 12.

We happened to know, AFTER THE FACT, that the other card was the same.

Put another way: Say I draw 4 cards out of the deck, and they are all tens. What are the odds the next card I draw will be a ten?

You are saying 1 out of 48 (52-4) and I am saying 0 chance (of the 12x4 remaining, none are tens.)

The "specificity" is accounted for.

As this is basic combinatorics, I think we can leave it off here.

But just so you know, if you click here I showed how to count all of the Gothic Chess positions before any one piece comes off of the board. That number is 32,099,674,107,692,140,366,789,953,222,888,490,987,180,838,400,000,000 which makes doing "card math" a piece of cake :)

I am currently writing a program to go through every possible set of two hands and checking for the same pair and three other (different) cards that are the same. When i'm done, we'll have the actual answer.

'

'
where basic algorithms are shared ... if you have access to the
Fred Fish Library of the Amiga Community, I'd recommend you
try out Mathlab from 1990 - it comes with an extra floppy full of old
Fortran sources - fancy implementations partly and, highly accurate
- fairly easy to transcode them to Pascal later ... ~*~

No more off-topic posts please. You have General Chat for it.